5.1.7 Gibbs Free Energy

What is Gibbs Free Energy?

The Gibbs free energy change ( $\Delta G$ ) combines the effects of enthalpy ( $\Delta H$ ) and entropy ( $\Delta S$ ) to predict the feasibility of a chemical reaction.

This is determined by the equation:

\Delta G = \Delta H - T\Delta S

where:

$\Delta G$ : Gibbs free energy change (kJ mol $⁻¹$ )
$\Delta H$ : Enthalpy change of the reaction (kJ mol $⁻¹$ )
$T$ : Temperature (in Kelvin, K)
$\Delta S$ : Entropy change of the reaction (J K $⁻¹$ mol $⁻¹$ )

For a reaction to be feasible, the value of $\Delta G$ must be zero or negative.

This means that the reaction is either spontaneous ( $\Delta G < 0$ ) or at equilibrium ( $\Delta G = 0$ ).

Balancing Entropy and Enthalpy

The feasibility of a reaction depends on both:

Enthalpy ( $\Delta H$ ): Reactions that release energy ( $\Delta H < 0$ ) favour feasibility.
Entropy ( $\Delta S$ ): Reactions that increase disorder ( $\Delta S > 0$ ) also favour feasibility.

The relationship between these two factors, along with temperature, dictates whether $\Delta G$ will be negative or positive.

Calculating Gibbs Free Energy

To calculate $\Delta G$ , use the given values for $\Delta H$ , $T$ , and $\Delta S$ in the equation:

\Delta G = \Delta H - T\Delta S

Remember to convert $\Delta S$ into kJ (if it's in J) by dividing by 1000 to match units with $\Delta H$

infoNote

Example Calculation

For the decomposition reaction:

\text{MgCO}_3(\text{s}) \rightarrow \text{MgO}(\text{s}) + \text{CO}_2(\text{g})

with values:

$\Delta H = +117 \, \text{kJ mol}^{-1}$
$\Delta S = +175 \, \text{J K}^{-1} \text{mol}^{-1}$
Temperature, $T = 298 \, \text{K}$

Step 1: Convert $\Delta S$ to kJ:

\Delta S = \frac{175 \, \text{J K}^{-1} \text{mol}^{-1}}{1000}

= 0.175 \, \text{kJ K}^{-1} \text{mol}^{-1}

Step 2: Substitute values into the equation:

\Delta G = 117 - (298 \times 0.175)

\Delta G = 117 - 52.15

= 64.85 \, \text{kJ mol}^{-1}

Conclusion:

Since $\Delta G$ is positive, the reaction is not feasible at 298 K.

Calculating $ΔS$ for vaporisation of Water

infoNote

Example: Vaporisation of Water During vaporisation, the water molecules shift from being relatively ordered in the liquid phase to highly disordered in the gaseous phase. This increase in disorder leads to a positive entropy change, which we can calculate using known values for enthalpy of vaporisation and temperature.

Formula:

The entropy change for a phase transition, such as vaporisation, can be calculated using the following formula:

\Delta S = \frac{\Delta H}{T}

where:

$\Delta S$ = entropy change for vaporisation (J K $⁻¹$ mol $⁻¹$ )
$\Delta H$ = enthalpy change for vaporisation (J mol $⁻¹$ )
$T$ = temperature at which the phase change occurs (in kelvin, K) For water, the enthalpy of vaporisation ( $\Delta H_{vap}$ ) at its boiling point (100°C or 373 K) is approximately 40.7 kJ mol $⁻¹$ .

Method:

Step 1: Convert the Enthalpy of vaporisation to Joules:

Since enthalpy values are often given in kJ, convert $\Delta H_{vap}$ to joules to match units for entropy:

\Delta H_{vap} = 40.7 \, \text{kJ mol}^{-1} = 40700 \, \text{J mol}^{-1}

Step 2: Set the Temperature in Kelvin: Use the boiling point of water in kelvin, which is 373 K:

T = 373 \, \text{K}

Step 3: Calculate $ΔS$ Using the Formula:

Substitute $\Delta H$ and $T$ into the equation:

\Delta S = \frac{40700 \, \text{J mol}^{-1}}{373 \, \text{K}}

\Delta S = 109.1 \, \text{J K}^{-1} \text{mol}^{-1}

So, the entropy change ( $\Delta S$ ) for vaporizing water at its boiling point is approximately 109.1 J K $⁻¹$ mol $⁻¹$ .

Determining Feasibility at Different Temperatures

By rearranging the Gibbs free energy equation, you can calculate the temperature at which $\Delta G = 0$ , which is the threshold for reaction feasibility:

T = \frac{\Delta H}{\Delta S}

infoNote

For example, using the following:

\Delta H = +117 \, \text{kJ mol}^{-1}

\Delta S = +0.175 \, \text{kJ K}^{-1} \text{mol}^{-1}

Then you can calculate the temperature:

T = \frac{117}{0.175} = 668.57 \, \text{K}

Thus, this reaction would become feasible at temperatures above 668.57 K.

Plotting $\Delta G$ Versus Temperature

Graphing $\Delta G$ against $T$ can help visualize how temperature impacts reaction feasibility:

A negative slope in the graph of $\Delta G$ versus $T$ indicates that the reaction may become feasible as temperature increases.
By analyzing the intercept, you can determine the temperature at which $\Delta G = 0$ , confirming the temperature at which the reaction becomes feasible.

Gibbs Free Energy Simplified Revision Notes for A-Level AQA Chemistry

5.1.7 Gibbs Free Energy

What is Gibbs Free Energy?

Balancing Entropy and Enthalpy

Calculating Gibbs Free Energy

Example Calculation

Calculating $ΔS$ for vaporisation of Water

Determining Feasibility at Different Temperatures

Plotting $\Delta G$ Versus Temperature

500K+ Students Use These Powerful Tools to Master Gibbs Free Energy For their A-Level Exams.

Other Revision Notes related to Gibbs Free Energy you should explore

Gibbs Free Energy Simplified Revision Notes for A-Level AQA Chemistry

5.1.7 Gibbs Free Energy

What is Gibbs Free Energy?

Balancing Entropy and Enthalpy

Calculating Gibbs Free Energy

Example Calculation

Calculating ΔSΔSΔS for vaporisation of Water

Determining Feasibility at Different Temperatures

Plotting ΔG\Delta GΔG Versus Temperature

500K+ Students Use These Powerful Tools to Master Gibbs Free Energy For their A-Level Exams.

Other Revision Notes related to Gibbs Free Energy you should explore

Calculating $ΔS$ for vaporisation of Water

Plotting $\Delta G$ Versus Temperature